$55$ gangsters shoot the nearest gangster to them, where all the distances between them are different. Prove that at least one gangster will survive.

The pair of gangsters with lowest pairwise distance will shoot each other. If some other gangster shoots any of those two, then there will be at most $52$ bullets aimed at the remaining $53$ gangsters, and hence at least one will survive.

If no other gangster shoots any of those two, the problem is reduced to the case of $53$ gangsters and we proceed by induction. At this point, it boils down to checking that the case for $3$ gangsters always ends up with one alive.

We will argue by contradiction. Assume everyone died.

Nobody can shoot the same guy twice since there are only 55 bullets and 55 people to kill.

So, WLOG$^{\ast}$,by renumbering $g_1$ shoots $g_2$, $g_2$ shoots $g_3$, ..., $g_{55}$ shoots $g_1$.

$^\ast$As N.S. noted there there can be different connected components, however since $55$ is odd one them must of length at least 3, or 1.

Denote $l_i$ the distance between $g_i$ and $g_{i+1}$, for $1\leq i\leq 54$ and $l_{55}$ the distance between $g_{55}$ and $g_1$.

Notice that $l_1>l_2$ since $g_2$ shoots $g_3$ and not $g_1$. Similarly $l_i>l_{i+1}$ which leads to a contradiction since $l_1>l_{54}>l_{55}$ and $l_{55}>l_1$.

Say $G$ is the set of gansters. Define $s:G\to G$ by saying that $g$ shoots $s(g)$. We need to show $g $ is not surjective.

Suppose $g$ is surjective. Since $G$ is finite it follows that $g$ is injective.

And since $s$ is a bijection, the simple inductive argument in the deleted answer works: Say $d(g_1,g_2)$ is the smallest distance. Them $s(g_1)=g_2$ and $s(g_2)=g_1$. Since $s$ is a bijection on $G$ it follows that $s$ maps $G'=G\setminus\{g_1,g_2\}$ to itself. (This last point was not clear in the deleted answer, which is presumably why it was deleted.) It follows by induction that there exists $g$ with $s(g)=g$.

Of course if each gangster literally shoots the nearest gagnster then $s(g)=g$ for all $g$. But clearly what was intended is that each gangster shoots the nearest other gangster, in which case $s(g)=g$ iis a contradiction.